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MOBILE GUARDS

3.1. INTRODUCTION

In this chapter we explore an interesting variant of the art gallery problemsuggested by Toussaint. Rather than modify the shape of the polygons as inthe previous chapter, we modify the power of the guard. Specifically, eachguard is permitted to "patrol" an interior line segment. Let s be a linesegment completely contained in the closed polygonal region P.s^P. Thenx e P is said to be seen by s, or is covered by s, if there is a point yes suchthat the line segment xy c P. Thus x is covered by the guard if x is visiblefrom some point along the guard's patrol path. This is the notion of weakvisibility from a line segment introduced in Avis and Toussaint (1981b)(strong visibility requires x to be seen from every point of s), a conceptfurther explored in Chapter 8.

The main reason that mobile guards are interesting is that they lead tosome clean theorems, some difficult theorems, and to interesting openproblems. Secondarily they connect to the important notion of edgevisibility, to be discussed further in Chapters 7 and 8. A covering by mobileguards induces a partition into edge-visible polygons.

We present two long proofs in this chapter. The first establishes that[n/4\ mobile guards are occasionally necessary and always sufficient tocover an n vertex polygon. The second proof, which is quite complex,establishes the equivalent result for orthogonal polygons: [(3«+4)/16jmobile guards are necessary and sufficient. This latter quantity may seemugly in comparison to the simpler fractions we have encountered so far, butthere is a clean logic behind it, as revealed in Table 3.1. Mobile guards aremore powerful than stationary guards: only 3/4's as many are needed, inboth general and orthogonal polygons—the second column is 3/4 times thefirst. Moreover, orthogonal polygons are 3/4's easier to cover than generalpolygons: the second row is 3/4 times the first. Thus orthogonal polygonsrequire about (3/4)2[n/3j mobile guards.

The first proof, presented in Section 3.2, is entirely combinatoric,following the outline of Chvatal's proof (Section 1.2.1). The second proof,

81

82

Guard —>Shape

General

Orthogonal

MOBILE GUARDS

Table 3.1

Stationary

[n/3\

[n/4\

Mobile

Ln/4JL(3n+4)/16j

presented in Section 3.3, is an instance where no reduction to combinatoricshas been discovered, and complex geometric reasoning seems necessary.The chapter closes with a discussion of related results.

3.2. GENERAL POLYGONS1

We first define various types of guards, both geometric and combinatorial.Three geometric mobile guards types with different degrees of patrolfreedom can be distinguished. An edge guard is an edge of P, including theendpoints. A diagonal guard is an edge or internal diagonal betweenvertices of P, again including the endpoints. A line guard is any linesegment wholly contained in P. (Recall that P is a closed region.)Geometric guards are said to cover the region they can see.

The combinatorial counterparts of these guards are obtained by defining aguard in a triangulation graph T of a polygon P to be a subset of the nodesof T. Then a vertex guard in T is a single node of T, an edge guard is a pairof nodes adjacent across an arc corresponding to an edge of P, and adiagonal guard is a pair of nodes adjacent across any arc of T. The analog ofcovering is domination: a collection of guards C = {gl) . . . , gk) is said todominate T if every triangular face of T has at least one of its three nodes insome gi e C.

The goal of this section is to prove that [n/4j combinatorial diagonalguards are sometimes necessary and always sufficient to dominate thetriangulation graph of a polygon with n > 4 vertices. It is clear that if atriangulation graph of a polygon can be dominated by k combinatorialvertex guards, then the polygon can be covered by k geometric vertexguards. The implication is that a proof of the sufficiency of a [«/4j ofcombinatorial diagonal guards in a triangulation graph establishes thesufficiency of the same number of geometric diagonal and line guards in apolygonal region.

Necessity is established by the generic example due to Toussaint shown inFig. 3.1: each 4 edge lobe requires its own diagonal guard.

1. An earlier version of this section appeared in O'Rourke (1983a), © 1983 D. Reidel.

3.2. GENERAL POLYGONS 83

Fig. 3.1. A polygon that requires [n/4j edge, diagonal, or line guards.

3.2.1. Sufficiency Proof

The proof is by induction and follows the main outlines of Chvatal'sinductive proof (and Honsberger's exposition (Honsberger 1976)). Beforecommencing the proof, it will be convenient to establish certain facts thatwill be used in various cases of the proof. The most important of theseconcerns "edge contractions." Let P be a polygon and T a triangulationgraph for P, and let e be an edge of P, and u and v the two nodes of Tcorresponding to the endpoints of e. The contraction of e is a transformationthat alters T by removing nodes u and v and replacing them with a newnode x adjacent to every node to which u or v was adjacent.2 Compare Figs.3.2a and 3.2d. Note that an edge contraction is a graph transformation, nota polygon transformation: the geometric equivalent ("squashing" thepolygon edge) could result in self-crossing polygons. Edge contractions arenevertheless useful because of the following lemma.

LEMMA 3.1. Let T be a triangulation graph of a polygon P, and 7" thegraph resulting from an edge contraction of 7. Then 7" is a triangulationgraph of some polygon P'.

Proof. We construct a figure with curved edges corresponding to 7", thenstraighten the edges to obtain P'.

Let Pt be the planar figure corresponding to the triangulation T, and let ebe the edge contracted and u and v its two endpoints in Pt. Let the verticesto which u and v are connected by diagonals and edges be y0, . . . , yt andz0, . . . , Zj, respectively, with y0 = v and z0 = u, and the remainder labeledaccording to their sorted angular order. See Fig. 3.2a. Note that y1 = z1 isthe apex of the triangle supported by e.

Now introduce a new vertex x on the interior of e, and connect the y andz vertices to x by the following procedure. Connect y1 to x; this can be donewithout crossing any diagonals because yx is the apex of a triangle on whosebase x lies. Remove the diagonal (u, yx). Connect y2 to x within the region

2. Harary calls this transformation an elementary contraction (Harary 1969).

84 MOBILE GUARDS

Fig. 3.2. If all the arcs in a triangulation graph (a) incident to u and v are made adjacent to x(b and c), the resulting graph may be deformed into a straight line graph (d).

bounded by (x,y1,y2, u); the line may need to be curved but again nocrossings are necessary. Remove the diagonal (u, y2). Continue in thismanner (see Fig. 3.2b) until all the v's have been connected to x. Thenapply a similar procedure to the z vertices. The result is a planar figurewhose connections are the same as those of 7". See Fig. 3.2c.

Finally, apply Fary's theorem (Giblin 1977): for any planar graph drawn


in the plane, perhaps with curved lines, there is a homeomorphism3 in theplane onto a straight-line graph such that vertices are mapped to verticesand edges to edges. Applying such a homeomorphism to the figureconstructed above yields P', a polygon that has V as one of itstriangulations. See Fig. 3.2d. •

The main use of this contraction result is the following.

LEMMA 3.2 Suppose that f(n) combinatorial diagonal guards are alwayssufficient to dominate any rc-node triangulation graph. Then if T is anarbitrary triangulation graph of polygon P with one vertex guard placed atany one of its n nodes, then an additional f{n -1) diagonal guards aresufficient to dominate T.

Proof. Let u be the node at which the one guard is placed, and let v be anode adjacent to u across an arc corresponding to an edge e of P. Edgecontract T across e, producing the graph T of n - 1 nodes. By Lemma 3.1T' is a triangulation graph, and so can be dominated by f(n - 1) diagonalguards. Let x be the node of T' that replaced u and v. Suppose that noguard is placed at x in the domination of T. Then the same guardplacements will dominate T, since the given guard at u dominates thetriangle supported by e, and the remaining triangles of T have dominatedcounterparts in T. Again compare Figs. 3.2a and 3.2d. If a guard is used atx in the domination of T, then this guard can be assigned to v in T, withthe remaining guards maintaining their position. Again every triangle of Tisdominated. •

We note in passing that the same lemma holds for other types of guards,but we will only need to use it with diagonal guards. Intuitively, one canview this lemma as saying that one edge can be "squashed" out for guardcoverage calculations if a guard is assigned to either of the edge's endpoints.

The next three lemmas establish special diagonal guard results for smalltriangulation graphs.

LEMMA 3.3. Every triangulation graph of a pentagon (n = 5) can bedominated by a single combinatorial diagonal guard with one endpoint atany selected node.

Proof. Let T be a triangulation graph of a pentagon, and let the selectednode be labeled 1. It is easy to show that there are only five distincttriangulations. In each case, a single combinatorial diagonal guard (pair ofadjacent nodes), with one end at node 1 can dominate the graph (see Fig.3.3). D

LEMMA 3.4 Every triangulation graph of a septagon (n - 7) can bedominated by a single combinatorial diagonal guard.

3 A homeomorphism is a continuous one-one onto mapping whose inverse is also continuous;intuitively it is a deformation without tearing or pasting—that is, it preserves topologicalproperties.

86

1 2 1 2 1 2 1 2 1 2

Fig. 3.3. A pentagon can be dominated by a single diagonal guard (shown dashed) with oneend at node 1.

Proof. Let T be a triangulation graph of a septagon, and let d be anarbitrary internal diagonal. This diagonal partitions the seven boundaryedges of T according to either 2 + 5 = 7 or 3 + 4 = 7; clearly the partition1 + 6 = 7 is not possible.

Case 1 (2 + 5 = 7). Let d = (1, 3). Then d supports another triangle T,either (1,3,4), (1,3,5), (1,3,6), or (1,3,7). Only two of these cases aredistinct.

Case la (T = (1, 3, 4)). Then (1, 4, 5, 6,7) is a pentagon (see Fig. 3.4a).By Lemma 3.3, this pentagon can be covered with a single diagonal guardwith one end of node 1. This guard dominates the entire graph.

Case lb (T = (1, 3, 5)). Choose diagonal (1,5) for the guard (see Fig.3.4b). Regardless of how the quadrilateral (1, 5, 6, 7) is triangulated, all ofT is dominated.

2 (3 + 4 = 7). Let d = (1, 4). Then both ways of triangulating thequadrilateral (1, 2, 3, 4) lead to situations equivalent to Case la above. •

LEMMA 3.5 Every triangulation graph of an enneagon (n = 9) can bedominated by two combinatorial diagonal guards such that one of theirendpoints coincides with any selected node.

Proof. Let T be a triangulation graph of an enneagon, let the selectednode be labeled 1, and let d be any internal diagonal with one end at 1. Thisdiagonal partitions the boundary edges of T according to either 2 + 7 = 9,3 + 6 = 9, or 4 + 5 = 9.

Case 1(2 + 7 = 9). Let d = (1, 3). The diagonal d supports another triangleT whose apex is at either 4, 5, 6, 7, 8, or 9. Only three of these cases aredistinct.

1 2 1 2

Fig. 3.4. A septagon can be dominated by a single diagonal guard.


I 2

Fig. 3.5. A enneagon can be dominated by two diagonal guards, with one of their ends atnode 1.

Case la (T = (1, 3, 4)). Dominate the septagon (1, 4, 5, 6, 7, 8, 9) with oneguard by Lemma 3.4, and use (1, 3) for the second guard (see Fig. 3.5a).

Case lb (T = (1, 3, 5)). Dominate the septagon (1, 3, 5, 6, 7, 8, 9) with oneguard by Lemma 3.4, and use (1, 3) for the second guard (see Fig. 3.5b).

Case lc ( r = (l, 3, 6)). Dominate the hexagon (1,2,3,4,5,6) with oneguard by Lemma 3.4, and dominate the pentagon (1,6,7,8,9) with oneguard whose endpoint is at 1 by Lemma 3.3 (see Fig. 3.5c).

Case 2 (3 + 6 = 9). Let d = (1, 4). If diagonal (1,3) is present, then wehave exactly Case la above. Otherwise diagonal (2, 4) is present, and oneguard along (1,2) together with a guard for the septagon as in Case lasuffices.

Case 3 (4 + 5 = 9). Let d = (1, 6). This is equivalent to Case lc above. •

Finally we must establish the existence of a special diagonal that willallow us to take the induction step, just as Lemma 1.1 did for Chvatal'sproof.

LEMMA 3.6. Let P be a polygon of n > 10 vertices, and T a triangulationgraph of P. There exists a diagonal d in T that partitions T into two pieces,one of which contains k = 5, 6, 7, or 8 arcs corresponding to edges of P.

Proof. Choose d to be a diagonal of T that separates off a minimumnumber of polygon edges that is at least 5. Let k>5 be this minimumnumber, and label the vertices 0 ,1 , . . . , n — 1 such that d is (0,k). See Fig.3.6. The diagonal d supports a triangle T whose apex is at t, 0 < t ̂ k. Sincek is minimal, t < 4 and k — t < 4. Adding these two inequalities yieldsA:<8. •

With the preceding lemmas available, the induction proof is a nearlystraightforward enumeration of cases.

THEOREM 3.1 [O'Rourke 1983]. Every triangulation graph T of a polygonof n >4 vertices can be dominated by [n/4\ combinatorial diagonal guards.

Proof. Lemmas 3.3, 3.4, and 3.5 establish the truth of the theorem for5 < n ^ 9, so assume that n ̂ 10, and that the theorem holds for all n' < n.Lemma 3.6 guarantees the existence of a diagonal d that partitions T into

tFig. 3.6. The diagonal d separates G into two pieces, one of which (Gj) shares 5<A:<8edges with G.

two graphs Tx and T2 where 7i contains k boundary edges of T with4 < k <: 8. Each value of k will be considered in turn.

Case 1 {k = 5 or 6). 71 has A: + 1 < 7 boundary edges including d. ByLemma 3.4, 7i can be dominated with a single diagonal guard. T2 hasn-k + l<n-5 + l = n-4 boundary edges including d, and by theinduction hypothesis, it can be dominated with |_("~4)/4j = \n/4\ — 1diagonal guards. Thus Tx and T2 together can be dominated by [n/4\diagonal guards.

Case 2 (k = 7). The presence of any of the diagonals (0, 6), (0, 5), (1,7),or (2, 7) would violate the minimality of k. Consequently, the triangle T in7i that is bounded by d is either (0,3,7) or (0,4,7); since these areequivalent cases, suppose that T is (0, 3, 7). The quadrilateral (0 ,1 , 2, 3) hastwo distinct triangulations. Each will be considered separately.

Case 2a ((1,3) is included.). Dominate the pentagon (3 ,4 ,5 ,6 ,7) withone diagonal guard with one end at node 3. This is possible by Lemma 3.3.This guard dominates all of 7i. Since T2 has n — 7 + l=n — 6 boundaryedges, it can be dominated by [(n - 6)/4j < [n/4\ - 1 diagonal guards bythe induction hypothesis. This yields a domination of T by [n/4j diagonalguards.

Case 2b ((0, 2) is included.). Form graph To by adjoining the two trianglesT = (0, 3, 7) and T = (0, 2, 3) to T2 (see Fig. 3.7). To has n -1 + 3 = n - 4edges, and so can be dominated by [(n - 4)/4j = [n/4\ - 1 diagonal guardsby the induction hypothesis. In such a domination, at least one of thevertices ofT' = (0, 2, 3) must be a diagonal guard endpoint. There are threepossibilities:

(0) If node 0 is a guard end, then To can be extended to include (0 ,1 , 2)without need of further guards.

(2) If node 2 is a guard end, then To can again be extended to include(0,1,2).


2" 4"

Fig. 3.7. Go is formed by adding T and T' to G2.

(3) If node 3 is a guard end, then there are three possible locations forthe other end of the guard. If the other end is at either node 0 or 2,then we fall into the two cases above. If the other end is at node 7,then replace the diagonal guard (3, 7) with (0, 7). Every triangle thatwas previously dominated is still dominated, and again To can beextended to included (0,1, 2).

Thus all but the pentagon (3,4,5,6,7) can be dominated with [n/4\ - 1diagonal guards, and the pentagon only requires a single diagonal guard byLemma 3.4, resulting in a total of [n/4\ diagonal guards for all of T.

Case 3 (k = 8). 7i has k + 1 = 9 boundary edges, and so by Lemma 3.5, itcan be dominated with two diagonal guards, one of whose endpoints is atnode 0. Now T2 has n — k + l = n -7 boundary edges. By Lemma 3.2, theone guard at node 0 permits the remainder of T2 to be dominated byf(n — 7 — 1) =/ (« - 8) diagonal guards, where the function /(«') specifies anumber of diagonal guards that are always sufficient to dominate atriangulation graph of n' nodes. By the induction hypothesis, /(«') =L«'/4j. Therefore, [(n — 8)/4j = [n/4j - 2 diagonal guards suffice to domi-nate T2. Together with the two allocated to Tly all of T is dominated by[n/4j diagonal guards. D

COROLLARY. Any polygon P of n >4 edges can be covered by [n/4jgeometric diagonal or line guards.

Proof. The diagonal guard result follows immediately from the theorem.Since diagonal guards are special cases of line guards, the same number ofthese more powerful guards clearly suffice. •

3.2.2. Edge Guards

The above proof depends on the fortunate identity between the number ofcombinatorial and geometric diagonal guards necessary and sufficient todominate and cover triangulation graphs and polygons, respectively. Thisidentity is not known to hold for edge guards, however. No polygons areknown to need more than [(n + l)/4j geometric edge guards (see Fig. 3.8),

90

7 I

Fig. 3.8. A polygon of seven edges that requires two edge guards.

but triangulation graphs exist that require [2n/7j = [«/3.5j combinatorialedge guards (see Fig. 3.9). Thus it appears that a different proof techniqueis required in this case.

19

Fig. 3.9. A triangulation graph that requires two edge guards per seven edges. The centraloctagon may be triangulated arbitrarily.

3.3. ORTHOGONAL POLYGONS

In this section we present Aggarwal's proof that [(3n + 4)/16j mobileguards are sufficient for covering an n vertex simple orthogonal polygon(Aggarwal 1984). The occasional necessity of this number of mobile guardsis established by a connected series of swastika-like polygons, as shown inFig. 3.10. The single swasktika shown in Fig. 3.10a with n = 20 requires fourguards, one per arm; note that (3 • 20 +4)/16 = 64/16 = 4. Merging two

3.3. ORTHOGONAL POLYGONS 91

Ln

Fig. 3.10. Polygons that require [(3n + 4)/16j mobile orthogonal guards: (a) n = 20 andg = 4; (b) n = 36 and g-1.

20-vertex swastika's together removes four vertices at the join, yieldingn = 36, as in Fig. 3.10b. This polygon requires seven guards, one for each ofthe six isolated arms, and one at the join; note that (3 • 36 + 4)/16 =112/16 = 7. Joining k swastikas results in an n = 16k + 4 vertex polygonsthat requires 3k + 1 guards; and note that [{3n + 4)/16j = 3k + 1. Thenecessity for other values of n is established by attaching a spiral of theappropriate number of edges to one arm of a swastika. Figure 3.11 showsthat a spiral addition of 6 edges requires one guard more than the swastika;a spiral addition of 12 edges requires two guards more. These are the criticaladditions; spirals with a different number of edges do not require a differentnumber of guards.

This establishes the necessity of [{3n + 4)/16j guards. We now turn tosufficiency. Aggarwal's proof is at least superficially similar in structure tothe proof for general polygons in the preceding section. The proof is byinduction. A small number of quadrilaterals are cut off from the givenpolygon, these small number covered separately, and the remainder of thepolygon handled recursively. The difficulties arise at the interface between

Fig. 3.11. Addition of a spiral establishes necessity for other values of n.

92 MOBILE GUARDS

the quadrilaterals cut off and the remainder. In the previous section,interfacing required choosing the diagonal guard with one end at theinterface, and applying the "edge-squashing" lemma (3.2) to reduce thenumber required in the remainder; in effect a guard is shared across theinterface. In Aggarwal's proof, the delicacy of the interface requires acomplex strategy to complete the induction proof.

Besides the increased complexity, the proof differs in two additionalaspects from that of Theorem 3.1. First, it uses geometric constructionsthroughout, as opposed to reducing the geometric problem to a purelycombinatorial one. It is unclear if this is essential; this point will be revisitedin Section 3.4. Second, the remainder of the polygon is often modified andneeds to be requadrilateralized. The proof of Theorem 3.1 maintained thesame triangulation throughout. The combination of these differences resultin a unique and complicated proof. It remains to be seen if a simplerapproach can establish the same result.

Before commencing with the details, it may be helpful to sketch the mainoutline of the proof. It will be shown below in Lemma 3.8 that there isalways a diagonal d in any quadrilateralization of an orthogonal polygonthat cuts off 2, 3, or 4 quadrilaterals. If four quadrilaterals are cut off by d,then properties of quadrilateralizations of orthogonal polygons permit onlytwo essentially different cases, and the induction carries through with a bitof sharing in the vicinity of d. If three quadrilaterals are cut off by d, thenthere are five distinct cases to handle, only one of which requires extensivesharing at the interface. Finally, if two quadrilaterals are cut off by d, thenthere are seven cases, most of which require sharing, some rathercomplicated. All the sharing is accomplished through one complex lemma(3.21). In all cases it will be shown that applying the induction hypothesis tothe remainder of the polygon, taking into account any interface sharing,results in [(3n + 4)/16j guards. We assume throughout that the polygon isin "general position" in that no two vertices can be connected by a verticalor horizontal line that does not intersect the boundary of the polygon.

We will first discuss structural properties of orthogonal polygons that willbe used throughout the remainder of the section. Then we will establish thelemmas used to share at the interface, and finally prove the theorem.

3.3.1. Properties of Orthogonal Polygons

We will conduct the argument in terms of the number of quadrilaterals q ina quadrilateralization of the polygon rather than in terms of the number ofvertices n. Our first two lemmas relate these quantities.

LEMMA 3.7. For any quadrilateralization of an orthogonal polygon of nvertices into q quadrilaterals, n = 2q + 2.

Proof. The sum of the interior angles of an orthogonal polygon of nvertices is 180(n - 2) degrees. But since there are q quadrilaterals, each of360 degrees, 360$ = 180(n - 2), or n = 2q + 2. •


Fig. 3.12. Diagonal d cuts off a minimum number of quadrilaterals that is at least 2.

The same lemma holds for any quadrilateralizable polygon, even those thatare not orthogonal.

Since q is fixed for any polygon, we will sometimes say "the number ofquadrilaterals in P" rather than "the number of quadrilaterals in anyquadrilateralization of P."

Applying this lemma to the sufficiency bound of [(3n + 4)/16j shows thatit is equivalent to [(3q + 5)/8j. It is in this form that the bound will appearthroughout the proof.

The following lemma is the equivalent of Lemma 3.6.

LEMMA 3.8. Let P be an orthogonal polygon and Q a quadri-lateralization of P. There exists a diagonal d in Q that partitions P into twopieces, one of which contains q = 2,3, or 4 quadrilaterals of Q.

Proof. Choose d to be a diagonal of Q that separates off a minimumnumber of quadrilaterals that is at least 2. Let q > 2 be this minimum. LetABCD be the quadrilateral supported by d = AB towards the piece with qquadrilaterals; see Fig. 3.12. The number of quadrilaterals in the BC, CD,and DA regions illustrated in the figure is each less than 2—that is, less thanor equal to 1—otherwise q would not be minimal. Therefore, q <4. •

It will often be useful to use the dual of a quadrilateralization. Let everyquadrilateral of a quadrilateralization Q be a node of a graph Q, where twonodes are adjacent in Q iff their corresponding quadrilaterals share adiagonal.4 The following is immediate (compare Lemma 1.3).

LEMMA 3.9. For any quadrilateralization Q of an orthogonal polygon,the dual Q is a tree with each node of degree no more than 4.

As an application of this observation, we can obtain an alternate proof ofLemma 3.8. Choose any root r for Q, and let x be a leaf at maximum

4 As mentioned in Chapter 1, this is the graph theoretic "weak dual," weak because no nodeis assigned to the exterior face.

94 MOBILE GUARDS

distance from r, and let y be the parent of x. Then all of the nodes adjacentto v not on the ry path must be leaf nodes; otherwise there would be a pathlonger than rx. Thus the diagonal of y that crosses the ry path cuts of 2, 3,or 4 quadrilaterals, depending on whether y is of degree 2, 3, or 4,respectively.

One of the main tools used throughout the proof is a cut, a toolpreviously used in Section 2.5. A cut L in an orthogonal polygon P is amaximal interior line segment in P (maximal in the sense that any linesegment properly containing L contains a point exterior to P) that containsan edge and a reflex vertex of P. L partitions P into two or three pieces,depending on whether it contains one or two reflex vertices respectively; seeFig. 3.13. In either case, the following holds.

LEMMA 3.10. The sum of the number of quadrilaterals in the piecesdefined by a cut L of P is equal to the number of quadrilaterals in P.

Proof. Suppose L partitions P into two pieces P1 and P2 as in Fig. 3.13a.Let P, Px, and P2 have n, n1, and n2 vertices and q, qlf and q2

quadrilaterals, respectively. Then n1 + n2 = n + 2, as L introduces one newvertex, counted in each of Px and P2. Lemma 3.7 shows that

n1 — 2 n2 — 2 n—2

If L partitions P into three pieces as in Fig. 3.13b, then L can be consideredas a combination of two "half cuts, each resolving just one reflex vertex.The first partitions P into two pieces, and the second partitions one of thepieces into two, resulting in three pieces. Applying the result justestablished for two pieces yields the lemma for three pieces. D

We now present a series of lemmas detailing the relationship between adiagonal of a quadrilateralization and the local structure of the polygon;henceforth "diagonal" means diagonal of a quadrilateralization. Recall thatthe orientation of an edge is horizontal or vertical. We will say that edges aand b are to the same side of d if they are in the same piece of P partitionedoff by d; note that a and b may be in opposite half-planes defined by d butstill to the same side.

I

i

1

a b

Fig. 3.13. A cut partitions a polygon into two (a) or three (b) pieces.


d/ b

Fig. 3.14. The five possible arrangements when a and b have opposite orientations. Thedotted lines represent possible orientations of the edges to the other side of d; the dashed linesindicate an added right angle that forms a subpolygon.

LEMMA 3.11. Let a and b be edges of P adjacent and to the same side ofa diagonal d. Then a and b have the same orientation.

Proof. Without loss of generality orient d with positive slope with thepolygon Px containing a and b below. Assume for contradiction that a and bhave different orientations. Then there are five distinct possible combina-tions of a and b: hanging up, down, left, or right from the endpoints of d, asshown in Fig. 3.14. The other three possible combinations force a and b tonot be to the same side of d. We can derive a contradiction in two ways.First note that in all five cases, addition of a right angle above d produces anew orthogonal polygon P'; perhaps it will be necessary to put this newangle on a different "level" as defined in Section 2.2, but this will not affectthe angle sums. Let qx be the number of quadrilaterals in P1. Then the sumof the internal angles of P' is 360q1 +180. But this implies that P' is notquadrilaterizable, in contradiction to Theorem 2.1.

For a second proof, recall Lubiw's scheme of assigning "types" to eachvertex of an orthogonal polygon such that they alternate type 1 and type 2in a traversal of the boundary (Section 2.4.2, especially Fig. 2.36). In all fivecases of Fig. 3.14, d connects two vertices of the same type. Thus the strictalternation is destroyed, and Px has an odd number of vertices. But thiscontradicts the assumption that Px is quadrilateralizable, since any polygonpartitioned into quadrilaterals must have an even number of vertices. •

Because no internal angle of a quadrilateral can be greater than 270°

96 MOBILE GUARDS

: b

Fig. 3.15. The four possible arrangements of a and b when d is in its standard orientation.The dotted lines indicate the possibilities for the edges adjacent to d and to a and b.

(since all are subangles of either 90° or 270°), only four configurations arepossible for d, a, and b, as illustrated in Fig. 3.15. We will raise thisobservation to a lemma for later reference.

LEMMA 3.12. The only configurations possible for a diagonal d and itstwo adjacent edges a and b to one side (perhaps after rotation and reflectionto orient d with positive slope) are those shown in Fig. 3.15.

Although not needed for the proof of the art gallery theorem, we nowturn our attention to characterizing the quadrilateral trees of orthogonalpolygons. This is accomplished by showing that Lemma 3.12 restricts theconfiguration possible for a quadrilateral of a specific degree to a finite setof possibilities, and that only certain configurations can "mate" with oneanother as adjacent quadrilaterals. We start by showing that a degree 4quadrilateral can have only one configuration. A configuration is defined bythe orientations of the edges of the polygon adjacent to each vertex of aquadrilateral, and the type (convex/reflex) of the vertices.

Let a reflex vertex whose exterior angle is in the first quadrant (betweenthe positive x and positive v axes) be called type 1, in the second quadrant(between the positive v and the negative x axes), type 2, and similarly fortype 3 and 4.

LEMMA 3.13. Let ABCD be a quadrilateral of degree 4 in Q for anyorthogonal polygon P. Then A, B, C, and D are each reflex vertices of P, oftypes 1, 2, 3, and 4 in counterclockwise order.

Proof. Assume to the contrary that at least A is convex. Without loss ofgenerality let A be a lower left corner as illustrated in Fig. 3.16a. Then Fig.3.15a shows that edge a' forces b and b' to have the orientations shown atB; b' forces c and c' as shown at C; and c' forces d and d' at D. But now d'lies inside ABCD, contradicting the assumption that ABCD is an internalquadrilateral.

3.3. ORTHOGONAL POLYGONS

| C

97

A Q

Fig. 3.16. If A is convex (a), d' is forced to be internal to ABCD; if A is reflex (b), the degree4 quadrilateral has a unique configuration.

Now let A be a type 3 reflex vertex. Following the same logic as aboveforces the configuration shown in Fig. 3.16b, establishing the lemma. •

The possible configurations proliferate for quadrilaterals of smallerdegree, but the proofs proceed the same way, repeatedly applying theconstraints imposed by Lemma 3.12, and will only be sketched.

LEMMA 3.14. A quadrilateral of degree three can have just one of thefour configurations shown in Fig. 3.17.

;'oe

- /\o

e-/

/+//e

c d

Fig. 3.17. The four configurations possible for a degree 3 quadrilateral.

98 MOBILE GUARDS

0 / 0 /

d e f

Fig. 3.18. The six configurations possible for a degree 2 quadrilateral.

Proof. Let e be the edge of the quadrilateral shared with the polygon. It iseasily shown using Lemma 3.12 that both endpoints of e cannot be convex.If one endpoint is convex and the other reflex, Fig. 3.17a is forced. If bothare reflex, three configurations are possible, shown in Figs. 3.17b-3.17d. •

The + , — , and 0 markings in Figs. 3.16 and 3.17 (and in the figures tofollow) will be explained later.

LEMMA 3.15. A quadrilateral of degree 2 can have just one of the sixconfigurations shown in Fig. 3.18.

Proof. If the two edges shared with the polygon are non-adjacent, thenthe three configurations shown in Figs. 3.18a-3.18c are possible. If theshared edges are adjacent, then the three configurations shown in Figs.3.18d-3.18f are possible. •

LEMMA 3.16. A quadrilateral of degree 1 can have just one of the twoconfigurations shown in Fig. 3.19.

This completes the classification of the possible configurations of thequadrilaterals in an orthogonal polygon. In order to study which configura-tions can mate with one another, we introduce the concept of "charge" on adiagonal. Let a and b be edges to the same side and adjacent to a diagonal d

a bFig. 3.19. The two configurations possible for a degree 1 quadrilateral.


a b c

Fig. 3.20. Definitions of the three diagonal charges.

of a quadrilateral. If a and b lie in the same half-plane determined by d,then we will say they have the same parity; otherwise they have oppositeparity. Thus in Figs. 3.15a and 3.15c, a and b have the same parity, and inFigs. 3.15b and 3.15d they have opposite parity. The charge on a diagonal dof a quadrilateral q, with respect to q, is 0 if the adjacent edges to bothsides of d have the same parity (Fig. 3.20a), + if the adjacent edges to the qside have the same parity, and the adjacent edges to the opposite side of dhave opposite parity (Fig. 3.20b), and — if the adjacent edges to the q sidehave opposite parity, and those to the opposite side of d have the sameparity (Fig. 3.20c). Note that charge is denned with respect to a qua-drilateral, so that each diagonal has a charge defined on either side.

LEMMA 3.17. The net charge on any diagonal in a quadrilateralization ofan orthogonal polygon must be zero: the charges must be 0/0, 4- / — , or- / + •

Proof. This is immediate from the definition of charge: a 0 charge on oneside is a 0 from the other side, and a + charge on one side is a — from theviewpoint of the other side. •

For the purpose of determining which configurations of quadrilaterals canmate with one another, each configuration can be reduced to a squaresymbol labeled with charges. The symbols corresponding to the configura-tions established in Lemmas 3.13-3.16 are displayed in Fig. 3.21 in the sameorder in which they appear in Figs. 3.16-3.19. We will refer to thesesymbols as, for example, [3b], meaning the b symbol for a degree 3quadrilateral as displayed in Fig. 3.21. All quadrilateral trees of orthogonalpolygons can be constructed by gluing these symbols together such that eachdiagonal is uncharged.

We may finally state and prove the characterization theorem.

THEOREM 3.2 [O'Rourke 1985]. A tree is a quadrilateral tree for asimple orthogonal polygon iff no node has degree greater than 4, and thetree contains no path connecting two degree 4 nodes by a sequence of zeroor more degree 3 nodes—that is, the path degree sequence (4 3* 4) does notoccur.

Proof. It is immediate that two degree 4 nodes cannot be adjacent, since

100 MOBILE GUARDS

0 0

Fig. 3.21. Symbols for all possible quadrilateral configurations. The numbers to the leftindicate the degree of the quadrilaterals, the letters below distinguish different configurations.

the symbol [4] has a negative charge on every diagonal. The degreesequence ( 4 3 3 3 . . . ) can be achieved by mating [4] with the + charge ofeither [3c] or [3d], and then mating - / + again with either [3c] or [3d], andso on. But it is clear that the last degree 3 quadrilateral in such a sequencehas two - diagonals free, neither of which can mate with [4]. Thus thedegree sequence (4 3 . . . 3 4) cannot occur in the quadrilateral tree of anyorthogonal polygon.

Now we show that any tree that does not contain a (4 3* 4) path can berealized as the quadrilateral tree of an orthogonal polygon, by assigningsquare symbols to each node such that all diagonals are uncharged. Assignto each degree 4 node the only choice, [4]. For each connected subtree Scomposed of degree 3 nodes, distinguish two cases. If S is adjacent to a [4],assign [3c] to each node in S, aligning the charges to balance. This will leaveonly - charges on the unmatched diagonals of S. If S is not adjacent to a[4], assign one of the leaves [3a], and all the other nodes of S [3c] as in thefirst case. Now there is one unmatched 0 diagonal, and the remainingunmatched diagonals are negatively charged. The important point is that nounmatched diagonal has a + charge. Next assign each degree 1 nodeadjacent to a negative diagonal [lb], and all others [la]. Note that a degree1 node will not be adjacent to a + charge by construction. Finally assign thedegree 2 nodes one of the symbols to cancel the charges appropriately.Since the only charge configuration not available with degree 2 nodes is onewith two negative diagonals, this will always be possible as long as a degree2 node doos not have to mate with two positive diagonals. But byconstruction, all free diagonals are either 0 or - . This completes theconstruction and the proof. •

The construction procedure is illustrated in Fig. 3.22. Figure 3.22a showsa tree that does not contain the forbidden degree sequence, and Fig. 3.22b


I 2

Ib

Ib

4

2b

2a

la

Ib

3c 3c

3c

Ib

la

3a

2c

3c

2b

Ib

Ib

la

Ib

\la

2

/

a

Ib

4

2b

' 3c

l V 3c

3 c /

\\\ /

Ib

Ib

2c

<.' 3o

Ib

la

\

Fig. 3.22. A non-forbidden tree (a), a selection of symbols matching the tree degrees (b), andan orthogonal polygon realizing the symbols (c).

shows the symbols assigned by the construction, glued together appropri-ately to cancel charges. Finally Fig. 3.22c shows an orthogonal polygon thatresults by replacing the symbols by their corresponding configurations. It isclear that there are many options in the transition from the symbols to theactual polygon, but the transition is always possible by adjusting the lengthsof the edges to avoid overlap, in a manner similar to the local scale changesused in Culberson and Rawlins (1985).

3.3.2. Sharing Lemmas

In this section we develop three "sharing lemmas" similar in spirit toLemma 3.2 in the proof for general polygons in Section 3.2. They all have

102 MOBILE GUARDS

Fig. 3.23. The partial shadow of a diagonal.

the following flavor: "Suppose the induction hypothesis holds, and we aregiven a polygon with one (or more) guards placed in particular locations'free.' Then an additional X guards suffice for total coverage." Here X willalways be just the right amount to establish the induction hypothesis. I amcalling these "sharing" lemmas because in effect they are sharing "frac-tional" guards across the induction dividing diagonal.

The induction hypothesis that is the premise of these lemmas is:

Induction Hypothesis (IH). Any orthogonal polygon with q'<qquadrilaterals may be covered with [(3q' + 5)/8\ mobile orthogonalguards.

First we present a specialized geometric lemma that will be needed in theproofs of the sharing lemmas. Let a and b be the two edges adjacent to andto the same side of a diagonal d, with the same parity. Thus we have eitherFig. 3.15a or 3.15c. These situations are clearly identical after rotation andreflection, and we will henceforth consider just Fig. 3.15a. In this situation,define the partial shadow of d to be the closed triangular region defined byd, a, and a vertical line through either x, the right endpoint of a, or throughthe vertex incident to d and b, whichever is leftmost. See Fig. 3.23.

LEMMA 3.18. The partial shadow of a diagonal in a quadrilateralizationof an orthogonal polygon is empty.

Proof. The partial shadow is only defined in the situation illustrated in Fig.3.23. Let A be the vertex incident to d and a as shown. Assume the shadowis not empty, and let e be the leftmost vertical edge in the shadow. Then Aand e must be part of a quadrilateral Q. But there is no vertex that canserve as the fourth for Q: it cannot lie to the right of e, for then Q would benon-convex; it cannot lie collinear with e, for then our general positionassumption is violated; nor can it lie to the left of e, since e is leftmost. •

The following lemma is almost the direct analog of Lemma 3.2.

LEMMA 3.19. If P is a polygon of q quadrilaterals with one guardplaced along a convex edge e (one whose endpoints A and B are bothconvex vertices), then assuming IH, P can be covered with an additionalL[3fal) + 5]/8j guards.

Proof. The proof is by induction on q. The lemma is clearly true whenq = L Assume it is true for q'<q. Let Q=ABCD be the quadrilateral


A e B A e B

Fig. 3.24. If ABCD has degree 1 and e is guarded, one quadrilateral may be removed.

containing e = AB. The proof proceeds by cases depending on the degree ofQ. If deg(Q) = 1, it follows easily; deg(Q) = 2 requires more work; anddeg(Q) = 3 is not possible.

Case (deg(Q) = 1). Either BC (or symmetrically DA) or CD is the soleinternal diagonal of Q. In either situation, illustrated in Fig. 3.24, a cutthrough C partitions P into a covered rectangle and a polygon of q — 1quadrilaterals by Lemma 3.10. Applying IH establishes the lemma.

Case (deg(Q) = 2). The internal diagonals of Q are either adjacent or not.

Case 2.1 (Non-adjacent Diagonals) The only situation possible is shown inFig. 3.25a, corresponding to Fig. 3.18a. The two cuts illustrated partition Pinto 3 pieces, a rectangle bound by the cuts, and two orthogonal polygons Px

and P2 of, say, qx and q2 quadrilaterals. By Lemma 3.10, qx + q2+ 1 = q.Now note that the guard along e is a guard between two convex vertices in

Fig. 3.25. If ABCD has degree 2 and e is guarded, one quadrilateral may be removed, eitherby induction (a), or by removal of a rectangle (b and c).

104 MOBILE GUARDS

each of Px and P2. Since q\<q and q2 < q, the induction hypothesis for thislemma applies. Therefore, P can be covered with

additional guards. Tedious analysis shows that this is less than or equal to[[3(q - 1) + 5]/8j, establishing the lemma.

Case 2.2 (Adjacent Diagonals). Only Fig. 3.18d is possible, which we willfurther partition into the two cases shown in Figs. 3.25b and 3.25c. Let BCand CD be the diagonals of Q; C must be above D. The two figures aredistinguished by whether x, the upper endpoint of the vertical edge incidentto B, is higher or lower than D. In the former case (Fig. 3.25b), a cutthrough D, and in the latter case (Fig. 3.25c), a cut through x, is guaranteedby the emptiness of the partial shadow of BC (Lemma 3.18) to partition Pinto a covered rectangle and a polygon of q — 1 quadrilaterals. Applying IHestablishes the lemma.

That the case deg(Q) = 3 is not possible is immediate from the possibleconfigurations shown in Fig. 3.17: e is not a convex edge in any of thepossible configurations. •

The next sharing lemma in effect "squashes out" two quadrilaterals.

LEMMA 3.20. If P is a polygon of q quadrilaterals with two guards placedon consecutive convex edges AB and BC, then assuming IH, P can becovered with an additional [[3(q — 2) + 5]/8j guards.

Proof. The proof is similar to the preceding one. The structural pos-sibilities are clearly the same as in that proof, but with BC here playing therole of AB there. Let Z, A, B, C, D, and E be consecutive vertices on theboundary of P. Let Q be the quadrilateral including BC; Q is notnecessarily ABCD.

Case 1 (deg(Q) = 1). Either Q = ABCD with AD the internal edge (Fig.3.26a), or Q = BCDE with BE the internal edge (Fig. 3.26b). In the firstinstance D is reflex, and a horizontal cut through it leaves a coveredrectangle and a polygon of q — 1 quadrilaterals that satisfies Lemma 3.19.Applying that lemma establishes the result. In the second instance, E isreflex, and a vertical cut leaves a covered rectangle and a polygon of q — 1

Fig. 3.26. If ABCD has degree 1 and AB and BC are guarded, one quadrilateral may beremoved.


A i—

Fig. 3.27. If ABCD has degree 2 and AB and BC are guarded, either induction applies (a),the situation is impossible (b), or a quadrilateral may be removed (c).

quadrilaterals that satisfies the induction hypothesis (and Lemma 3.19). Inall cases, then, the result holds.

Case 2 (deg(Q) = 2). The same two cases apply as in Lemma 3.19.

Case 2.1 (Non-adjacent Diagonals.) The situations must be as in Fig.3.27a. The two cuts Lx and L2 partition P into a covered rectangle, apolygon Pl with qx quadrilaterals that satisfies the induction hypothesis, anda polygon P2 of q2 quadrilaterals that satisfies Lemma 3.19, whereqx + q2 + l=q. Applying both results yields coverage with

additional guards. A tedious analysis reveals this to be no larger thanL[3fo-2) + 5]/8j for 4 > 2.

Case 2.2 (Adjacent Diagonals) A cannot be a vertex of Q: Figs. 3.18d,3.18e, and 3.18f do not permit three consecutive convex vertices. ThereforeQ = BCDX, corresponding to Fig. 3.18d, with D reflex. Now if X is aboveA, as in Fig. 3.27b, Z is in the partial shadow of BX, a contradiction. So weare left with the situation shown in Fig. 3.27c. A cut through D establishesthe result as in Case 1 (compare Fig. 3.26a). •

The last sharing lemma is the most complex. It takes the form: if certainsharing conditions hold, then the remainder of the polygon needs one full

106 MOBILE GUARDS

guard less than [(3q + 5)/8j. The sharing conditions are rather complicated,but essentially the idea is to place two guards crossing each otherorthogonally such that the previous two lemmas apply to the pieces of theresulting partition.

LEMMA 3.21. If P is a polygon of q quadrilaterals with a guard placedalong a maximal segment Lx that contains a polygon edge that is situated inP such that

(a) there are two cuts orthogonal to Lx that partition off rectanglestouching (and thereby covered by) Lx, and

(b) Lx cuts the remainder (P with the two rectangles removed) into oneor two (i.e., not three) pieces.

then assuming IH, P can be covered with an additional [(3q + 5)/8j — 1guards.

Proof. Note that Lx is not necessarily a cut, but could be a convex edge.Let Px and P2 be the two pieces separated by Lx after removal of the tworectangles; P2 may be empty. The premise of the lemma is a bit ungainly,but is composed to have two geometric consequences:

(1) If P2 is not empty, there is at least one reflex vertex on Lx in Px U P2.(2) Lx lies on a convex edge of both P1 and P2.

We first support these claims. If Lx is a cut, it partitions P into two or threepieces. If Lx cuts P into three pieces, the premise can only be satisfied if thethird piece is composed of one or both of the rectangles cut off, disallowingFig. 3.28a for example. If Lx cuts P into two pieces, then if the second pieceis composed of one or both of the cut off rectangles, then P2 is empty, as in

Fig. 3.28. The cut in (a) does not satisfy the conditions of the lemma; in (b), P2 is empty; in(c) and (d), Lx contains a reflex vertex.


Fig. 3.28b, for example. Let V be the reflex vertex on Lx (the one in Px U P2

in case there are two reflex vertices on Lx). Then either V is in Px U P2, as inFig. 3.28c, or there is at least one other reflex vertex on Lx introduced bythe orthogonal cuts, as in Fig. 3.28d. Finally, since Lx is a supporting linefor both Px and P2, it constitutes a convex edge in each.

Let Px and P2 have qx and q2 quadrilaterals. Then qx + q2 = q -2 byLemma 3.10. We now apply the previous two sharing lemmas to establishthe claim.

By property (2), Lemma 3.19 applies to both Px and P2, resulting incomplete coverage with

additional guards. A tedious case analysis shows that this quantity is nogreater than [(3q + 5)/8j - 1 for all possible mod 8 residues of qt and q2

except in the single case when both g1 = 2(mod8) and <72 = 2(mod8). Wenow concentrate on this "hard" case. Note that P2 cannot be empty in thiscase.

Introduce a cut L2 orthogonal to Lx through the reflex vertex Vguaranteed by property (1), and place a guard along L2. L2 must partitionone of Px or P2, say P2, into two pieces P2 and P2, with q2 and q2

quadrilaterals; it may or may not partition P1} as illustrated in Fig. 3.29. IfL2 partitions Plf call the pieces P[ and P" with q[ and q'{ quadrilaterals.Now although q[ + q'{ = qx by Lemma 3.10, q2 + q'2' = q2+l, since Lx

already resolved the reflex vertex V.Note that the conditions for the application of Lemma 3.20 hold for both

P2 and P2; Lx and L2 both lie on convex edges in each. Therefore, all of P2

. P|" JL, P2 ^ \

1-2

P2 j

Fig. 3.29. The cut L2 may (a) or may not (b) partition Px.

108 MOBILE GUARDS

can be covered with

g2 = [[3(q2 - 2) + 5]/8j + [[3(q'i - 2) + 5]/sJ (1)

guards. If L2 does not partition Px, then we apply Lemma 3.20 to Pi tocover it with

g1=[[3(q1-2) + 5]/s\ (2)

guards. If L2 does partition P1} then Lemma 3.20 can be applied to P[ aboveL2, and Lemma 3.19 to P'[ below L2, resulting in

g[ = [[3(q[ - 2) + 5]/8J + [[3{q'[ - 1) + 5]/8j (3)

guards. Using the special case assumption that qx = 8k\ + 2, (2) yieldsg\ = 3k1} and a case analysis and q[ + q'i = qi shows that (3) impliesg[ < 3kx. Therefore, 3kx guards suffice for Px in either case. The assumptionq2 = 8A:2 + 2 and q2 + q2 = qi + 1 leads to (1) to g2 ^ 3k2. Thus a total of3(k1 + k2) guards suffice. Finally, q = qx + q2 = 8(kx + k2) + 6 implies thatL(3g + 5)/8j - 2 = 3{kx + k2), which together with the 1 guard placed on L2,establishes the lemma. •

3.3.3. Proof of Orthogonal Polygon Theorem

We have finally assembled enough lemmas to prove the main theorem.

THEOREM 3.3 [Aggarwal 1984]. [{3q + 5)/8j = [(3/i + 4)/16j mobileguards are sufficient to cover any orthogonal polygon P of q quadrilateralsand n vertices.

Proof. The proof is by induction on q. If q < 2, then 1 guard clearlysuffices. Assume now the induction hypothesis IH. Fix an arbitraryquadrilateralization of P. Lemma 3.8 established that there is a diagonal dthat cuts off a minimal number A: of 2, 3, or 4 quadrilaterals. Theseconstitute the three cases of the proof, which we consider in reverse order.

Case k = 4. Recall from the proof of Lemma 3.8 (see Fig. 3.12) that dmust be a diagonal of a degree 4 quadrilateral Q, say Q = ABCD withd = DA. Lemma 3.13 shows that A, B, C, and D must all be reflex vertices.Let A be left of and lower than D, which can be achieved without loss ofgenerality by rotation and reflection. We can distinguish three cases, onlytwo of which are real possibilities, depending on the horizontal sorting of B,C, and D. We will use the notation X < Y to mean that point X is strictlyleft of point Y.

Subcase (C<D (Fig. 3.30a)). This case violates Lemma 3.11, as comple-tion of the polygon between B and C as illustrated demonstrates.

Subcase (C>D and B <D (Fig. 3.30b)). Place a guard on the vertical cutLj through D as illustrated. Then this cut satisfies the conditions of Lemma


b

A/

D

/

Z_

\ c

B

c

A

D

XC

/B

Fig. 3.30. If ABCD has degree 4, then either the situation is impossible (a), or Lemma 3.21applies (b and c).

3.21, with P2 empty. Applying that lemma yields coverage of P1 with[(3q + 5)/8\—l guards, which, together with the guard along Llt

establishes the theorem.

Subcase (C>D and B>D (Fig. 3.30c)).Lx the vertical cut through D.

Again Lemma 3.21 applies with

Case k = 3. The proof of Lemma 3.8 shows that d is a diagonal of a degree3 quadrilateral Q. Let Q = ABCD with d = DA. Orient d as in Fig. 3.31a,and assume without loss of generality that the edges of P' adjacent to d arehorizontal, with A reflex. (Figure 3.17 shows that at most one of A, B, C, Dis convex, so one end of d is always reflex. If the other end is convex, dangles away from the reflex vertex as in Fig. 3.31a; if the other end is reflex,then either d angles away as in Fig. 3.31a, or it will after reflection in the xaxis.) We distinguish five cases, depending mainly on which edge of Q is apolygon edge. In each case, Lemma 3.21 is invoked.

Subcase (BC is a polygon edge.) BC must be horizontal and below A,

110 MOBILE GUARDS

D

/

B

Fig. 3.31. The configuration when k — 3: the dotted edges in (a) represent the two possibleorientations of the polygon edge at D. If BC is a polygon edge, Lemma 3.21 applies.

otherwise either Q is non-convex or Lemma 3.11 is violated. Consequentlythe parity of the horizontal edges adjacent to d in P' is the same, and thesituation is as illustrated in Fig. 3.31b. A horizontal cut through B satisfiesthe conditions of Lemma 3.21 (with P2 empty), and the theorem follows byplacing a guard along the cut and applying Lemma 3.21.

Subcase (CD is a polygon edge). CD must be vertical to satisfy Lemma3.11, and B must be left and below C since Q is convex. Regardless of thevertical placements of A, B, and C, a vertical cut through B satisfies Lemma3.21. Figure 3.32 shows that in each of the three possible vertical sortings(A, C, B), (C, A, B), and (C, B, A), in a, b, c respectively, the theoremfollows by placing a guard along the cut and applying Lemma 3.21.

Subcase (AB is a polygon edge). Distinguish further subcases, dependingon the location of C with respect to B and D.

•

/B

Fig. 3.32. If CD is a polygon edge, Lemma 3.21 applies.


Fig. 3.33. If AB is a polygon edge, Lemma 3.21 applies in all cases.

Subsubcase (C is below B and left of D (Fig. 3.33a).). Let Lx be themaximal vertical segment containing D. This satisfies Lemma 3.21, regard-less of whether or not D is reflex.

Let Lx be theSubsubcase (C is below B and right of D (Fig. 3.33b).).vertical edge containing C. This satisfies Lemma 3.21.

Subsubase (C is above B and left of D). This case violates Lemma 3.11and so is not possible.

Subsubcase (C is above B and right of D (Fig. 3.33c).). Let Lx be themaximal vertical segment containing D. This satisfies Lemma 3.21.

Case k = 2. Although this case is simplest in some sense, it requires themost extensive sharing, since so little is cut off by d. Fortunately, all thesharing is concentrated into Lemma 3.21. We partition the problem into twosubcases, depending on whether the edges adjacent to d in P' have the sameor opposite parity. Let d = DA as usual, and let A be below D so that A isalways reflex.

Subcase (Same Parity). Let d be oriented as in Fig. 3.34a. Place a guardalong the vertical edge through D if D is convex (Figs. 3.34b and 3.34c), oralong the first vertical edge hit by a horizontal cut through A (Fig. 3.34d).In all cases, Lemma 3.21 applies, with P2 empty.

Subcase (Opposite Parity). Orient d as in Fig. 3.35a. That both A and D

A

D

c dFig. 3.34. When the edges adjacent to d have the same parity, Lemma 3.21 applies in allcases.

Fig. 3.35. When the edges adjacent to d have opposite parity, Lemma 3.21 applies in allcases.


are reflex with their adjacent edges oriented as shown can be seen byexamination of Fig. 3.18. Of the four vertices in the chain counterclockwisebetween A and D, exactly one is reflex. If the first or second (counterclock-wise from A) is reflex (Figs. 3.35b and 3.35c), a vertical cut Lx through Dsatisfies the conditions of Lemma 3.21. If the third or fourth vertex from Ais reflex (Figs. 3.35d and 3.35e), a vertical cut through A satisfies Lemma3.21. In all cases, placing a guard along Lx and applying Lemma 3.21 yieldscoverage by [(3q + 5)/8j - 1 + 1 guards, establishing the theorem.

D

1/

y A

C

X

C D

Fig. 3.36. An "execution" of the proof of Theorem 3.3 and the lemmas it invokes. The finalguard placement is shown in (g).

114 MOBILE GUARDS

We have exhausted all possibilities, and therefore the theorem is estab-lished. •

The proof just presented is constructive, and therefore can be convertedto an algorithm. The algorithm is highly inefficient, however, sincerequadrilateralization is implicitly required at almost every step. It will helpunderstanding the proof if we step through a small example, tracking theproof through the various lemmas and "executing" them as procedures.

Consider the polygon shown in Fig. 3.36a. It has n = 26 vertices andq = 12 quadrilaterals. The theorem then says that six guards suffice; actuallyfour suffice in this case. Using the quadrilateralization in Fig. 3.36a, d is adiagonal that cuts off a minimum number k of quadrilaterals; in this case,k = 2. Following the theorem, the k = 2 case (opposite parity: Fig. 3.35breflected) invokes Lemma 3.21. In our particular case, the cut Lx and theabutting rectangles are shown. Lx partitions P into pieces with qx = 2 andq2 = 10 quadrilaterals. This is the hard case of the lemma, and requires asecond cut L2 shown. Two non-trivial pieces remain, and for both Lemma3.21 invokes Lemma 3.20, because there are guards on two consecutiveconvex edges (Fig. 3.36b). Both pieces fall under the same case of Lemma3.20 (deg(Q) = 1: Fig. 3.26a reflected), both introducing a cut and invokingLemma 3.19 for a guard along a single convex edge. Figure 3.36c shows thesmaller piece. Lemma 3.19 makes a cut (following Fig. 3.24b) and appliesthe IH, which in this case is trivial since the remaining piece is a rectangle,which is assigned its own guard. Figure 3.36d shows the larger piece. AgainLemma 3.19 cuts and applies IH to the polygon shown in Fig. 3.36e. We arenow back at the "top level" in the main theorem. In the quadrilateralizationshown, d cuts off a minimum k = 2 quadrilaterals. The case here is the sameparity one (Fig. 3.34c), and introduces a guard along the vertical edgeshown. The top remainder is handled by Lemma 3.19, because the guardforms a convex edge. Lemma 3.19 then (unnecessarily in this case) invokesIH again, this time at the basis, and a guard is assigned to the rectangle inFig. 3.36f. The resulting five guards assigned are shown in Fig. 3.36g.

3.4. Discussion

The guards used for the general polygon theorem (Theorem 3.1) arecombinatorial: visibility is needed only at the two vertices at the endpointsof diagonals. The guards used for the orthogonal polygon theorem(Theorem 3.3) are geometric: visibility is required throughout their length.The guards used in the two theorems differ in other respects. Severalfeatures of the orthogonal guards are:

(1) Visibility is required throughout the length of the guard.(2) Guards are oriented horizontal or vertical only.(3) Each guard can be chosen to include an edge of the polygon.(4) Visibility is only required orthogonal to the guard.(5) The patrols of two guards may pass through one another.

3.4. DISCUSSION 115

These were not conditions imposed on the problem, but rather those that"fell out" of Aggarwal's proof. It would be interesting to disallow the fifthcondition above: do not permit the lines of two guards to cross. But themost interesting question concerning these qualifications on guard "power"is whether (1) is necessary: can the same result be achieved withcombinatorial guards, as in the general polygon case?

Aggarwal has proven several other results on mobile guards (Aggarwal1984). The most important is that for quadrilaterizable polygons—that is,those that can be partitioned into convex quadrilaterals, [n/5\ guards arenecessary and sufficient. Since [n/5\ > [(3n +4)/16j for all n>20, thisresult does not contradict Theorem 3.3. Despite Theorem 3.2, whichcharacterizes the quadrilateral trees of orthogonal polygons, it remains anopen problem to characterize those polygons that are quadrilateralizable.Aggarwal's proof of the [n/5\ result differs in two ways from the proof ofTheorem 3.3: first, it is entirely combinatorial, and second, it is muchlonger: at one point 93 separate cases are considered! Several other of hismobile guard results for specialized polygons will be discussed in the nextchapter.

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